Sat Physics

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(2) About Peterson’s To succeed on your lifelong educational journey, you will need accurate, dependable, and practical tools and resources. That is why Peterson’s is everywhere education happens. Because whenever and however you need education content delivered, you can rely on Peterson’s to provide the information, know-how, and guidance to help you reach your goals. Tools to match the right students with the right school. It’s here. Personalized resources and expert guidance. It’s here. Comprehensive and dependable education content—delivered whenever and however you need it. It’s all here.. Artwork by Timothy J. Finley Editorial Development: Sonya Kapoor Turner Special thanks to my wife, Faye, and Mrs. Cathy Walton. For more information, contact Peterson’s, 2000 Lenox Drive, Lawrenceville, NJ 08648; 800-338-3282; or find us on the World Wide Web at www.petersons.com/about. COPYRIGHT. © 2002 Peterson’s. Previous edition. © 2001.. ALL RIGHTS RESERVED. No part of this work covered by the copyright herein may be reproduced or used in any form or by any means—graphic, electronic, or mechanical, including photocopying, recording, taping, Web distribution, or information storage and retrieval systems—without the prior written permission of the publisher. For permission to use material from this text or product, complete the Permission Request Form at http://www.petersons.com/permissions.. ISBN 0-7689-0960-0 Printed in the United States of America 10 9 8 7 6 5 4 3 2 1. 04 03 02.

(3) CONTENTS. Red Alert .....................................................1 About the Book .......................................................... 1 About the Physics Test ............................................... 2 Scoring .............................................................................. 3 Taking the Test .................................................................. 3. SAT II Physics Study Plans ......................................... 4 Preliminary Preparations .................................................. 5 The 18-week Plan .............................................................. 6 The 9-Week Plan ............................................................... 8 Panic Plan .......................................................................... 9. Key Formulas and Equations ................................... 11 Units and Conversions ............................................. 12. Diagnostic Test ...........................................13 Answers and Explanations ...................................... 31. Chapter 1: Preliminary Concepts .................45 Simple Equations and Algebra ................................. 47 Graphs ...................................................................... 48 Right Triangles ......................................................... 52 Units and Conversions ............................................. 56 Scalars and Vectors ................................................... 57 Chapter Summary .................................................... 67. Chapter 2: Mechanics .................................69 Statics ...................................................................... 71.

(4) CONTENTS. Torques ..................................................................... 77 Kinematics ............................................................... 80 Motion in Two Dimensions...................................... 87 Newton’s Laws of Motion ........................................ 90 Work and Energy ..................................................... 96 Momentum............................................................. 100 Circular and Rotary ................................................ 103 Chapter Summary .................................................. 109. Chapter 3: Waves ....................................113 Wave Properties ..................................................... 115 Chapter Summary .................................................. 135. Chapter 4: Heat and Thermodynamics ......................................137 Temperature ........................................................... 139 Thermal Properties of Matter ................................ 140 Thermodynamics ................................................... 149 Chapter Summary ................................................. 160. Chapter 5: Electricity and Electromagnetism .....................................161 Electrostatics .......................................................... 163 Electric Fields ......................................................... 166 Electric Circuits ...................................................... 168 Magnets and Magnetic Fields ................................ 179 Chapter Summary .................................................. 191.

(5) CONTENTS. Chapter 6: Modern Physics .......................193 Particulate Theory of Light .................................... 195 Photoelectric Effect ................................................ 198 Relativity ................................................................. 201 Chapter Summary .................................................. 211. Chapter 7: The Atom.................................213 The Atom ................................................................ 215 Radioactivity ........................................................... 224 Particles .................................................................. 232 Chapter Summary .................................................. 234. Practice Test 1 ..........................................237 Practice Test 2 ..........................................271 Practice Test 3 ..........................................305 Practice Test 4 ..........................................337.

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(7) RED ALERT ABOUT THE BOOK Now that you have decided to tackle the SAT II Physics Test, you need to develop a plan to maximize your chances of achieving a high score. The reason you want a high score is simple. The higher your score, the better you look to the admissions officers at the college of your choice. The SAT II Physics Test is designed to measure the knowledge and achievement of high school students who have completed a college preparatory physics class. This book will help you review for the SAT II Physics Test. If you have not taken a Physics class yet, it may be advisable to wait until after you complete one before you progress further. This book is arranged in several sections, beginning with a diagnostic test. The purpose of this test is to identify your strengths and weaknesses. It will also help to familiarize you with the testing method used by the College Board. Do not skip the diagnostic test! Find a quiet, comfortable spot, turn off the TV, computer, and music, and try to take the entire diagnostic test within the time allotted. If you run out of time, mark on your answer sheet where you ran out of time. Then keep going until you complete the test. After finishing the test, you deserve a break, so take one. But be sure to come back after no more than 15 minutes and check your answers. Then go back, reread the questions you missed, and try to figure out what went wrong. Did you jump to a conclusion, misread the question, stop at the first correct answer (sometimes there are multiple correct answers to one question), or just not know the material? None of these reasons is cause for serious concern right now, because every reason mentioned can be overcome. In fact, as you work through this book, each of these will be addressed. Once the diagnostic test is done it’s time to settle down to your preparation for the real SAT II Physics Test. Move on to the Physics review and study each section as if you had to learn the material all over again. Those sections you know well will go by smoothly and help you to build confidence, while your review of unfamiliar terms and concepts will allow you to learn it quickly and effectively.. Peterson’s: www.petersons.com. RED. 1. ALERT.

(8) RED ALERT. When you have finished your review, take the practice tests. Once again, you should try to set up a test-like situation. Make sure all the electronic toys are out of the way (besides, the College Board doesn’t allow them), time yourself, and do your best. After you check each test, read the explanations for those questions you missed, and for additional reinforcement, reread the Review Section that covers the material in the question.. ABOUT THE PHYSICS TEST The SAT II Physics Test is a 75-question multiple-choice test. The test covers the following topics: • • • • • •. Mechanics Heat/Kinetic Theory/Thermodynamics Waves Electricity and Magnetism Modern Physics Miscellaneous Topics (may include measurement, math, and history). Nearly 40 percent of the SAT II Physics Test will test your knowledge of Mechanics. Electricity and Magnetism also make up a significant proportion of the material, accounting for about 25 percent of the questions. Approximately 20 percent of the questions deal with waves, and the remaining 15 percent covers the other topics listed above. Because the material covered in high school physics courses can vary widely, you will probably find some questions on the test that cover topics with which you are unfamiliar. This may be impossible to avoid entirely; however, solid preparation and review of the concepts and information covered in this book will go a long way to helping you navigate unfamiliar territory. The testing procedures for the SAT II Physics Test are similar to those you already know from your high school classroom. A few questions will be simple recall; about half of the questions will require you to be able to apply a physical concept for a given situation; and the more difficult questions will require that you be able to apply multiple concepts to multiple relationships. In addition to a college preparatory physics course, you will also need a solid working knowledge of algebra and trigonometry, as well as laboratory experience. Familiarity with the metric system is critical. You ar e not per mitted to use a calculator dur ing the test. are permitted during The calculations focus on simple arithmetic and will not require a calculator.. RED. 2. ALERT. Peterson’s SAT II Success: Physics.

(9) RED ALERT. SCORING Your score on the SAT II Physics Test will be reported on a scale of 200–800. Each question you answer correctly counts as one point. Each incorrect answer counts as 1 4 point against you. Unanswered questions do not count for or against you. Clearly, the more incorrect answers you can eliminate from a set of choices, the better your chances of finding or guessing the correct answer.. TAKING THE TEST You will be given one hour (60 minutes) to complete seventy-five questions. Because of the limited amount of time you have for each question, it is vital that you adopt a test-taking strategy and stick to it. There are several things that you can do to give yourself the best possible chance to score well on the test. Set up a study program for yourself and faithfully follow it. Do not skip any part(s) of the program. 1. When you take the diagnostic and practice tests, make sure to familiarize yourself thoroughly with the test directions, as they are patterned after the actual directions on the SAT II Physics Test you will be taking. Becoming familiar with the directions saves you time in re-reading them on test day. 2. Quickly read the entire test question by question. Answer immediately the questions you know or can do as you read through the test the first time; skip those questions you can’t answer right away. 3. Once you’ve gone through the entire test, go back to the beginning and work on the questions you skipped. Look for answers to eliminate. The more answers you can eliminate, the better your chances to recognize or guess the correct answer. 4. If you finish the test before time is up, go back and look over your answers. Only change an answer if you are absolutely sure you have the incorrect answer selected. 5. Get a good rest the night before the test. A primary reason for poor test performance is lack of sleep.. Peterson’s: www.petersons.com. RED. 3. ALERT.

(10) RED ALERT. SAT II PHYSICS STUDY PLANS You already know the importance of a study plan when preparing for this test. Needless to say, the amount of time you have before you’re due to take the test has a lot to do with which of the following plans you’ll select for your exam preparation. Those of you who are not taking the SAT II Physics Test in the near future should follow our leisurely 18-week plan. This plan is favored because it gives you plenty of time to thoroughly prepare, review all required concepts, and restudy the material you find challenging. The next option is the 9-week plan, which calls for a more concentrated effort on your part. You will have to pay more attention to your diagnostic and practice test results in this plan, as those questions you miss become indicators for the material to which you should pay extra attention. Finally, the last method is the Panic Plan. This plan is for you if you’ve got hardly any time to prepare but still want to do your best, like when you find out your college of choice wants to see your SAT II Physics Test score and the next test administration is in just a few weeks! Using this plan you will spend as much time as you have available preparing for the SAT II Physics Test by using this book to line you up for the test.. RED. 4. ALERT. Peterson’s SAT II Success: Physics.

(11) RED ALERT. PRELIMINARY PREPARATIONS Browse through the book. Look at the chapter summaries. Take the Diagnostic Test. Follow the instructions for taking the test and then check your answers. After you have taken the test and graded it, be sure to read all the explanations for the answers. Pay particular attention to the explanations for the questions you missed. Try to identify the reasons why you answered a question incorrectly. Was it carelessness on your part? You would be surprised at the number of students who miss a question they knew by inadvertently marking the wrong choice. Perhaps you misread the question, or maybe you were hurrying too much. Look through the answers and explanations for the questions you missed. Be sure you understand why the correct answer is correct. Write down the numbers of the questions you missed or would like to understand a little better. They are your benchmark questions. If you miss a lot of the questions about magnetism or thermodynamics, for example, this tells you where to concentrate your efforts. Identifying these weak areas is especially important if you are following the Panic Plan and will be helpful no matter how long you have to prepare. Sometimes students say the test questions are tricky, but the students who know the material are difficult to fool and quickly eliminate the “tricky” answers. This is not to say that they don’t have to think about what they are doing. Some of the questions are truly challenging and will require your best effort, so prepare yourself well. Start each chapter by reading the summary at the end, which lists everything in the chapter. As you read through a chapter, work out the problems on your own when you come to them. When you reach the werful end of a chapter, read the summary again. Repetition is a po pow learning tool.. Peterson’s: www.petersons.com. RED. 5. ALERT.

(12) RED ALERT. THE 18-WEEK PLAN Week 1. Diagnostic Test Review the answers to the Diagnostic Test. Review Chapter 1: Preliminary Concepts. Week 2. Chapter 2 2:: Mechanics Statics Kinematics Dynamics. Week 3. Chapter 2 2:: Mechanics Work and Energy Momentum Circular and Rotary Motion. Week 4. Chapter 3: Waves Wave Properties Reflection Refraction. Week 5. Chapter 3 3:: Waves Polarization of Light Interference Diffraction. Week 6. Chapter 4 mod ynamics 4:: Heat and Ther Thermod modynamics Thermal Properties of Matter Kinetic Molecular Theory Gases. Week 7. Chapter 4 mod ynamics 4:: Heat and Ther Thermod modynamics Laws of Thermodynamics Heat Engines Calorimetry. Week 8. Chapter 5 5:: Electricity and Electromagnetism Coulombs Law Electric Fields and Potential DC Circuits. Week 9. Chapter 5 5:: Electricity and Electromagnetism Magnets and Magnetic Fields. RED. 6. ALERT. Peterson’s SAT II Success: Physics.

(13) RED ALERT. Week 10. Chapter 6 6:: Modern Physics Quantum Mechanics Work Function. Week 11. Chapter 6 6:: Modern Physics Relativity Heisenberg Compton deBroglia. Week 12. Chapter 7 7:: The Atom The Nucleus Atomic Spectra Bohr’s Atom. Week 13. Chapter 7 7:: The Atom Nuclear Reactions, Equations, and Radiation Fission and Fusion Binding Energy and Mass Defect. Week 14. Pr actice Test 1 and Pr actice Test 2 Practice Practice Review Answers and Explanations Reread all the chapter summaries. Week 15 Review the material from the questions you missed on Practice Test 1 and Practice Test 2 Reread the chapter summaries. Week 16. Pr actice Test 3 and Pr actice Test 4 Practice Practice Review Answers and Explanations Reread all the chapter summaries. Week 17 Review the material from the questions you missed on Practice Test 3 and Practice Test 4 Reread the chapter summaries. Week 18 Start the week by reading all of the chapter summaries. Go over all of the test questions you missed on the Diagnostic Test and the four practice tests. Review the material you are not sure you have mastered! Think positive. You are ready!. Peterson’s: www.petersons.com. RED. 7. ALERT.

(14) RED ALERT. THE 9-WEEK PLAN Week 1. Diagnostic Test Chapter 1 eliminar y Concepts 1:: Pr Preliminar eliminary. Week 2. Chapter 2 2:: Mechanics. Week 3. Chapter 3 3:: Waves. Week 4. Chapter 4 mod ynamics modynamics 4:: Heat and Ther Thermod. Week 5. Chapter 5 5:: Electricity and Electromagnetism. Week 6. Chapter 6 6:: Modern Physics. Week 7. Chapter 7 7:: The Atom. Week 8. Pr actice Tests 1 and 2 Practice Review the answers to the questions you missed. Week 9. Pr actice Tests 3 and 4 Practice Review the answers to the questions you missed. Reread the chapter summaries, paying particular attention to any material you missed previously. RED. 8. ALERT. Peterson’s SAT II Success: Physics.

(15) RED ALERT. PANIC PLAN This last plan is for the group of students (and I hope it is small) who for whatever reason don’t have much time to prepare for the SAT II Physics Test. Perhaps you have just completed a physics class and you don’t think you need to spend much time in preparation, or maybe you have just decided to take the test. Only you know how much study time you have and how much energy you are willing to devote to preparation. Try to use both your time and energy wisely. The list below will help you to prepare for the test in whatever time you have available. By all means try to do everything on the list. If that is not a possibility, the most important items are first. Do them! In fact do as much as you can. • Read the chapter summaries, which will help you renew your basic physics knowledge. • Take the Diagnostic Test, and review the questions you miss. This gives you an idea of your needs. • Take Practice Test 1, and go over any questions you miss. • Take Practice Test 2, and go over the questions you miss. • Take Practice Test 3, and go over the questions you miss. • Take Practice Test 4, and go over the questions you miss. • Reread the chapter summaries Good luck!. Peterson’s: www.petersons.com. RED. 9. ALERT.

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(17) KEY FORMULAS AND EQUATIONS ∆y Slope = ∆x 2. 2. r =y +x. nλ = d sinq. 1 S = Vo t + at 2 2. PV = n r t P1 V1 = P2 V2. 2. S=. y sin q = r. 1 2 gt 2. V1 V2 = T1 T2. F = ma. P1 P2 = T1 T2. Wt = mg. x cos q = r. Work = F • S • cos q PE = mgh. tan q =. y x. KE= l/2 mv. Power =. ∑F=0 ∑ Fx = 0 ∑ Fy = 0. P = mv. ∑ Fz = 0. Ft = m∆V. Τ = F≥. Fc =. ∑Τ=0 ∑ +Τ = ∑ −Τ. Speed (U). distance d = time t. V=. V f + V0 2. ∆V = Vf – V0 ∆t = t f –to. a=. a=. ∆V ∆t. V f − V0 t f − t0. Vf = V0 + at Vf2 = Vo2 +2as. Peterson’s: www.petersons.com. ( P1 )(V1 ) ( P2 )(V2 ) = T1 T2. 2. ∆Q = ∆U + ∆W. work time. Q = cm∆T. F=K. (q1 )(q2 ) r2. F = EQ. mV 2 r. V = IR. Rt = R1 + R2 + R3 + R…. (m )(m ) F =G 1 2 2 r θradians =. S V= t. 1 1 1 1 = + + Rt R1 R2 R 3. arclength s = radius r. P = VI F = B⊥ IL. stan = qr. F = B⊥ qV. Vtan = Wr. EMF = BLV. atan = αr. Vs N s = Vp N p. V= λf. f =. Relativistic Factor. 1 Τ. E = hf φ = hf. hi q = h0 p. ∆ E = E2 –E1. 1 1 1 = + f p q n1sin q1 = n2 sin q2 (Snell’s Law). 11.   1   v 2  1− (c) .      . E = mc2. Activity =. T1/2 =. ∆N ∆t. .693 λ. ∆N = λN o ∆t.

(18) UNITS AND CONVERSIONS. SI Units length. meter. m. mass. kilogram. kg. time. second. s. electric current. ampere. A. temperature. Kelvin. K. Metric Prefixes T. tera. 1 × 1012. 10 12. G. giga. 1 × 109. 10 9. M. mega. 1 × 106. 10 6. hK. hectokilo. 1 × 105. 10 5. ma. myria. 1 × 104. 10 4. K. kilo. 1 × 103. 10 3. h. hecto. 1 × 102. 10 2. d. deka. 1 × 101. 10 1. Basic Unit. 1 meter – 1 gram – 1 liter. d c m. deci centi milli. 1 × 10–1 –2 1 × 10 1 × 10–3. 10 –1 –2 10 10 –3. dm. decimilli. 1 × 10–4. 10 –4. cm. centimilli. 1 × 10–5. 10 –5. u. micro. 1 × 10–6. 10 –6. n. nano. 1 × 10–9. 10 –9. p. pico. 1 × 10–12. 10 –12. 12. Peterson’s SAT II Success: Physics.

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(21) PHYSICS TEST. DIAGNOSTIC TEST PHYSICS TEST While you have taken many standardized tests and know to blacken completely the ovals on the answer sheets and to erase completely any errors, the instructions for the SAT II Physics Test differ in an important way from the directions for other standardized tests. You need to indicate on the answer key which test you are taking. The instructions on the answer sheet will tell you to fill out the top portion of the answer sheet exactly as shown. 1. Print PHYSICS on the line under the words Subject Test (print). 2. In the shaded box labeled Test Code fill in four ovals: —Fill in oval 1 in the row labeled V. —Fill in oval 6 in the row labeled W. —Fill in oval 3 in the row labeled X. —Fill in oval C in the row labeled Y. —Leave the ovals in row Q blank. Test Code. V W X Q. 1. 1. 2. 3. 4. 5. 6. 7. 8. 9. 1. 2. 3. 4. 5. 6. 7. 8. 9. 2. 3. 4. 5. Y. A. B. C. D. 1. 2. 3. 4. 5. 6. 7. 8. 9. Subject Test (print). Physics E. There are two additional questions that you will be asked to answer. One is “How many semesters of physics have you taken in high school?” The other question lists courses and asks you to mark those that you have taken. You will be told which ovals to fill in for each question. The College Board is collecting statistical information. If you choose to answer, you will use the key that is provided and blacken the appropriate ovals in row Q. You may also choose not to answer, and that will not affect your grade. When everyone has completed filling in this portion of the answer sheet, the supervisor will tell you to turn the page and begin. The answer sheet has 100 numbered ovals, but there are only approximately 75 multiple-choice questions on the test, so be sure to use only ovals 1 to 75 (or however many questions there are) to record your answers.. Peterson’s: www.petersons.com. 15.

(22) DIAGNOSTIC TEST. PHYSICS TEST Par artt A. Directions: Each of the sets of lettered choices below refers to the questions and/or statements that follow. Select the lettered choice that is the best answer to each question and fill in the corresponding oval on the answer sheet. In each set, each choice may be used once, more than once, or not at all.. Use the choices below to answer the questions that follow. The questions relate to a hydrogen electron located at E-3.. Questions 1-3 relate to the following chart, which is a partial energy level diagram for the hydrogen electron.. −.54 eV. n=5. −.85eV. n=4. −1.51eV. n=3. −3.4 eV. n=2. −13.6eV. n =1. A) B) C) D) E). ±..66eV ±.. 966eV ±1.. 89eV ±10.. 2eV ±12.. 09eV. 1. What is the emission energy when the electron falls to E-2 from E-3? 2. What is the absorbed energy when the electron jumps to E-5 from E-3? 3. What is the emission energy when the electron falls to E-1 fromE-3?. 16. Peterson’s SAT II Success: Physics.

(23) PHYSICS TEST. PHYSICS TEST— Contin ued TEST—Contin Continued Questions 4–6 relate to the changes that could be made in the following scenario and the results such changes would produce.. Questions 7–9 relate to the following scenario. A dog walks 120 m due east before turning and running 45 m west. He then turns and trots 40 m due north. After completing his journey, he is 85 m northeast of his home. When he hears his master call him, he runs directly home.. A pendulum swings at a rate of .75 vibrations/ second. A) B) C) D) E). Mass of the bob was increased Length of the pendulum was increased Mass of the bob was decreased Length of the pendulum was decreased Displacement from zero was increased. A) B) C) D) E). 4. The period of the vibrations would increase because…. The eastward leg The westward leg The northward leg The distance from home The distance to home. 7. Which part of the trip is a negative vector?. 5. The frequency of the vibrations would decrease because…. 8. Which part of the trip is an equilibrant? 9. Which part of the trip is the longest vector?. 6. The velocity of the pendulum would increase because…. Peterson’s: www.petersons.com. 17.

(24) DIAGNOSTIC TEST. PHYSICS TEST— Contin ued TEST—Contin Continued Questions 10–12 relate to the following. A) B) C) D) E). The choices below give a description of the quantities listed above. Match the statement below with the quantity it describes above.. Frequency Amplitude Wavelength Velocity Period. 10. The number of wave crests passing a given point per unit of time. 11. The distance between two points or two consecutive waves. 12. The product of the frequency and the wavelength.. Par artt B Directions: Each question or statement below is followed by five possible answers. In each case, select the best possible answer and fill in the corresponding oval on the answer sheet.. 13. A wooden crate is pushed across a concrete floor at 5 m/s and released. It slides to a stop after moving a short distance. The same crate is filled until it weighs twice as much as it did previously and again slid across the floor at 5 m/s and released. The stopping distance for the crate will be (A). 1 as far. 4. (B). 1 as far. 2. (C) (D) (E). the same distance. twice as far. four times as far.. 14. A team of skydivers jumps from a plane and holds hands to form a flower-like design. As the skydivers begin their free fall, their velocity increases and their (A) (B) (C) (D) (E). 18. acceleration increases. acceleration decreases. acceleration is constant. acceleration is zero. air resistance is reduced.. Peterson’s SAT II Success: Physics.

(25) PHYSICS TEST. PHYSICS TEST— Contin ued TEST—Contin Continued 18. The catcher prepares to receive a pitch from the pitcher. As the ball reaches and makes contact with his glove, the catcher pulls his hand backward. This action reduces the impact of the ball on the catcher’s hand because (A) the energy absorbed by his hand is reduced. (B) the momentum of the pitch is reduced. (C) the time of impact is increased. (D) the time of impact is reduced. (E) the force exerted on his hand remains the same.. 15. A professional golfer drives a golf ball 230 meters down the fairway. When the club head strikes the golf ball (A) the impact force on the golf ball is greatest (B) the impact force on the club head is greatest (C) the impact force is the same for both. (D) the impact force has no effect on the club (E) the impact force has no effect on the ball 16. When a woman pushes on her grocery cart, the woman moves because of (A) the force the woman exerts on the grocery cart. (B) the force the grocery cart exerts on the woman. (C) the force the woman exerts on the ground. (D) the force the ground exerts on the woman. (E) the force the grocery cart exerts on the ground. 17.. 19. A 12,500 kg boxcar rolling through a freight yard has a velocity of 1 m/s when it strikes another boxcar of the same mass that is at rest. Both cars stick together and continue to roll down the track with a momentum of (A) 0 kg • m/s (B) 3125 kg • m/s (C) 6250 kg • m/s (D) 12,500 kg • m/s (E) 25,000 kg • m/s. During a company picnic, 6 accounting department workers participate in a tug of war with 6 sales force personnel. Each team pulls on the rope with 1200N of force. What is the tension in the rope? (A) 2400N (B) 1200N (C) 600N (D) 200N (E) 100N. Peterson’s: www.petersons.com. 20. A 750 g peregrine falcon dives straight down towards a 400 g pigeon, which is flying level to the ground. Just before the falcon makes impact its velocity is 35 m/s. The velocity of the falcon and the pigeon in its talons immediately after impact is most nearly (A) 35 m/s (B) 31.95 m/s (C) 28.9 m/s (D) 25.85 m/s (E) 22.8 m/s. 19.

(26) DIAGNOSTIC TEST. PHYSICS TEST— Contin ued TEST—Contin Continued 21. A father holds his child on his shoulders during a parade. The father does no work during the parade because (A) no force acts on the child. (B) the momentum of the child is constant. (C) the potential energy of the child is gravitational. (D) the child’s kinetic energy is constant. (E) the child’s distance from the ground remains the same.. 24. During a laboratory experiment, a 19.6N pile driver is dropped 2 m on to the head of a nail, which is driven 2.45 cm into a wood board. The frictional force exerted by the wood on the nail is (A) 96.04N (B) 165N (C) 1600N (D) 1960N (E) 3200N 25. For question 24 above, what is the magnitude of the acceleration of the pile driver while it drives the nail into the board? (A) –165 m/s2 (B) –800 m/s2 (C) –1600 m/s2 (D) –2000 m/s2 (E) –3200 m/s2. 22. Golden Glove boxers, who are amateurs, use larger, more padded gloves than professional boxers use. The amateur boxers are more protected from injury because (A) the larger glove exerts a larger impulse on the boxer. (B) the larger glove exerts a larger force on the boxer. (C) the larger glove exerts more energy on the boxer. (D) the larger glove increases time of impact on the boxer. (E) the larger glove increases the power exerted on the boxer.. 26.. 23. The driver of an automobile traveling at 80 km/hr locks his brakes and skids to a stop in order to avoid hitting a deer in the road. If the driver had been traveling at 40 km/hr, how much faster would he have stopped? (A) 4 times the distance (B) 2 times the distance (C). A child is swinging on a swing set. As the child reaches the lowest point in her swing (A) the tension in the rope is equal to her weight. (B) the tension in the rope supplies a centrifugal force. (C) her kinetic energy is at maximum. (D) her tangential acceleration equals gravity. (E) her angular velocity is minimum.. 27. A bicycle wheel spins on its axis at a constant rate but has not yet made a complete rotation. Which of the following statements is correct? (A) The angular displacement is zero. (B) The linear displacement is zero. (C) The angular acceleration is zero. (D) The angular velocity is zero. (E) None of these is zero.. 1 the distance 2. (D) 1 the distance. 4 (E) Not enough information to tell. 20. Peterson’s SAT II Success: Physics.

(27) PHYSICS TEST. PHYSICS TEST— Contin ued TEST—Contin Continued 31. A person who normally weighs 900N at sea level climbs to the top of Mt. Everest. While on top of Mt. Everest that person will weigh (A) zero. (B) approximately 900N. (C) considerably less than 900N but more than zero. (D) considerably more than 900N. (E) Need more information.. 1 somer2 sault. During his dive he uses the tuck position so that he will have (A) larger angular momentum. (B) smaller angular momentum. (C) larger rotational rate. (D) smaller rotational rate. (E) longer time in the air.. 28. An Olympic diver performs a 3. 29. While riding on a merry-go-round, you decide to move from a position close to the center to a position on the outside rim of the merry-go-round. After you have changed position, which of the following has remained the same? (A) Tangential acceleration (B) Centripetal force (C) Angular displacement (D) Tangential velocity (E) Tangential displacement. 32. If a shaft were drilled through the center of the earth and all you had to do was step into the shaft to “fall” to the other side, and a 800N person took the trip, her weight at the exact time she passed through the exact center of the earth would be (A) zero. (B) 800N. (C) less than 800N but more than zero. (D) more than 800N. (E) Need more information.. 30. The International Space Station is currently under construction. Eventually, simulated earth gravity may become a reality on the space station. What would the gravitational field through the central axis be like under these conditions? (A) zero (B). 1g 4. (C). 1g 2. (D). 3g 4. 33. A 750N person stands on a scale while holding a briefcase inside a freely falling elevator. Which of the following is true? (A) If the briefcase were released it would rise to the ceiling. (B) The person’s acceleration is zero. (C) The person’s attraction toward the earth is zero. (D) The person’s apparent weight is zero. (E) If the briefcase were released it would fall to the floor.. (E) 1 g. Peterson’s: www.petersons.com. 21.

(28) DIAGNOSTIC TEST. PHYSICS TEST— Contin ued TEST—Contin Continued 34. The time it takes a satellite to make one orbit around the earth depends on the satellite’s (A) acceleration. (B) weight. (C) direction of rotation. (D) distance from earth. (E) launch speed.. 38. The volume of an ideal gas is reduced to half its original volume. The density of the gas (A) remains the same. (B) is halved. (C) is doubled. (D) is tripled. (E) is quadrupled.. 35. Geosynchronous satellites remain over the same spot on the earth’s surface because they (A) orbit the earth every 24 hours. (B) are in polar orbits. (C) rotate opposite the earth’s rotational direction. (D) have a varying orbital height. (E) use terrain reading technology to remain on station.. 39. Refrigerators and freezers perform their functions by (A) converting hot air to cold air. (B) keeping hot air out with cold air pressure. (C) removing heat from inside themselves. (D) blowing cold inside them. (E) producing cold air. 40. An empty soda can with a few ml of water inside is heated to steaming and quickly inverted into an ice water bath. The can is instantly crushed because (A) energy in the can is lost. (B) water vapor condenses leaving a vacuum, which sucks the can in. (C) water vapor condenses and outside air pressure crushes the can. (D) the cold water shrinks the hot can. (E) water pressure in the ice bath crushes. the can. 36. Rutherford’s results in his famous gold foil experiment proved that atoms (A) are mostly space. (B) are in continuous motion. (C) have negative orbitals. (D) have diffuse charge distribution. (E) have dense crystalline structure. 37. A hanging weight stretches a spring 8 cm. If the weight is doubled and the spring constant is not exceeded, how much will the spring stretch? (A) 4 cm (B) 8 cm (C) 12 cm (D) 16 cm (E) 20 cm. 41. The first law of thermodynamics is a restatement of (A) (B) (C) (D) (E). 22. Guy-Lassac’s Law. the principle of entropy. the principle of enthalpy. conservation of energy. Avogadro’s hypothesis.. Peterson’s SAT II Success: Physics.

(29) PHYSICS TEST. PHYSICS TEST— Contin ued TEST—Contin Continued 42. If 360 g of water at 95°C is mixed with 275 g of water at 10°C, what is the resulting temperature of the water? (A) 37°C (B) 49°C (C) 58°C (D) 70°C (E) 82°C. Questions 45–47 refer to the waves shown below. Each wave is moving in the direction shown by the arrow.. Select the graph that answers the questions below.. 43. While anchored at sea, a captain notices the wave peaks are separated by 16 m and occur at a rate of 1 wave every 2 seconds. What is the velocity of these waves? (A) 4 m/s (B) 8 m/s (C) 16 m/s (D) 32 m/s (E) 64 m/s. 45. Which set of pulses will soon show constructive interference? 46. Which set of pulses has already been through interference? 47. Which set of pulses will soon show destructive interference?. 44. Some opera singers are able to use their voice to shatter a crystal glass. They can do this because of (A) acoustic reflection. (B) multiple echoes. (C) interference. (D) resonance. (E) beats.. Peterson’s: www.petersons.com. 48. A rider on a subway train hears the engineer blow the train whistle. A moment later she hears an answering whistle from an approaching train. Why does the whistle she hears from the approaching train change pitch? (A) The frequency of the waves of the approaching train’s whistle is decreasing. (B) The frequency of the waves of the approaching train’s whistle is increasing. (C) The loudness of the waves of the approaching train is decreasing. (D) The loudness of the waves of the approaching train is increasing. (E) The echoes of the two trains’ whistles are combining.. 23.

(30) DIAGNOSTIC TEST. PHYSICS TEST— Contin ued TEST—Contin Continued 53. Which of the following best describes the electric field about a positive point charge? (A) The field strengthens as the distance from the point charge increases. (B) The field is a constant throughout space. (C) The field weakens as the distance from the point change increases. (D) The field is oriented toward the point charge. (E) The field cannot be determined.. 49. Two charged objects are moved 50% closer to one another. Which statement about the electric force between the objects is true? (A) The force between them doubles. (B) The force between them halves. (C) The force between them remains the same. (D) The force between them reverses. (E) The force between them operates in the same direction. 50. The distribution of the charge density on the surface of a conducting solid depends upon (A) the density of the conductor. (B) the shape of the conductor. (C) the size of the conductor. (D) the age of the conductor. (E) the substance of the conductor.. 54. A 6 volt battery is connected across a resistor, and a current of 1.5 A flows in the resistor. What is the value of the resistor? (A) 2Ω (B) 4Ω (C) 6Ω (D) 8Ω (E) 10Ω. 51. A pair of point charges, which have charges of –3 micro coulombs and –4 micro coulombs, is separated by 2 cm. What is the value of the force between them? (A) 600N (B) 300N (C) 540N (D) 270N (E) 400N. 55. As a battery ages, its internal resistance increases. This causes the current in the external circuit to (A) remain the same. (B) polarize. (C) reverse direction. (D) increase. (E) decrease.. 52. At what point between a pair of charged parallel plates will the electric field be strongest? (A) It is strongest between the plates. (B) It is strongest near the positive plate. (C) It is strongest near the negative plate. (D) The field is constant between the plates. (E) The field is variable, therefore the strong point also varies.. 24. Peterson’s SAT II Success: Physics.

(31) PHYSICS TEST. PHYSICS TEST— Contin ued TEST—Contin Continued 59. A charged particle moving through a magnetic field will experience the largest force when (A) moving with the field. (B) moving against the field. (C) moving at a 45° angle to the field. (D) moving at a 90° angle to the field. (E) the particle will not be affected.. Questions 56–58 refer to the circuit below.. 60. Which of the following is caused completely or in some part by magnetic lines of force? (A) The picture on a computer screen (B) Radio reception interference (C) Aurora Borealis (D) V.H.S. films (E) All of these. 56. What is the resistance in the parallel circuit above between points B and C? (A) 2Ω (B) 4Ω (C) 6Ω (D) 8Ω (E) 10Ω 57. The (A) (B) (C) (D) (E). 61. Electrical energy is converted into mechanical energy by which of the following? (A) Magnet (B) Transformer (C) Motor (D) Generator (E) Battery. current in the circuit is 1A 2A 3A 9A 18 A. 62. A transformer contains 4000 turns on its primary side and 500 turns on the secondary side. If the input voltage is 240 V, calculate the voltage output of the secondary side.. 58. What is the voltage change between point B and point C? (A) (B) (C) (D) (E). (A) (B) (C) (D) (E). 1V 2V 3V 9V 18 V. Peterson’s: www.petersons.com. 25. 15 V 30 V 60 V 120 V 240 V.

(32) DIAGNOSTIC TEST. PHYSICS TEST— Contin ued TEST—Contin Continued 63. While standing in front of a plane flat mirror and looking at yourself, you raise your right hand. Which is the best description of the image you see? (A) Erect and enlarged (B) Erect and reduced (C) Erect and reversed (D) Inverted and reversed (E) Inverted and reduced. Questions 66 and 67 refer to the drawing below. The drawing represents the results in a Young’s double slit interference pattern experiment.. 64. Which of the following occurs when light passes into a clear glass cube? (A) The light’s wavelength changes. (B) The light’s frequency changes. (C) The light’s speed changes. (D) The light is polarized. (E) Both A and C. 66. Which letter represents the wavelength of the light? 67. Which letter represents the zeroth order fringe?. 65. A light ray is moving parallel to the principal axis of a concave mirror, which it strikes. How will the light ray be reflected? (A) Back upon itself (B) Through the focal point (C) Through the radius of curvature. 68. What does the photoelectric effect demonstrate? (A) The particulate nature of light (B) The wave nature of light (C) The diffuse reflection of light (D) The total internal reflection of light (E) The polarization of light. 1 f 2 (E) Through a point equal to 2r. (D) Through a point equal to. 69. An electric current is applied to a gas discharge tube, causing it to glow. When the discharge is viewed through a spectroscope, which type of spectrum is seen? (A) Monochromatic (B) Continuous (C) Line absorption (D) Line spectra (E) Polarimeter. 26. Peterson’s SAT II Success: Physics.

(33) PHYSICS TEST. PHYSICS TEST— Contin ued TEST—Contin Continued 73. What is the major product of the fission reaction of 235 U? The basic fission reaction is. 70. Einstein based his theory of relativity on the postulate that (A) photoelectrons absorb and emit photons. (B) the velocity of light is the same for all observers. (C) mass and energy are intra-convertible. (D) the universe is an entropic system. (E) energy and momentum are conserved.. 1 235 1 141 U+ p= Ba + 3 p + ? 0 92 0 56 (A) (B) (C) (D) (E). 71. DeBroglie theorized that all moving objects emit waves (matter waves) based on their h ö æ momentum ç λ = ÷ . Accordingly, as your mv ø è team’s defensive end, it is your job to stop the other team’s 250 pound fullback. If you could hear the fullback’s matter waves and you listened as the opposing fullback received the ball and accelerated toward you, what sound would you hear? (A) An increase in loudness and an increase in frequency (B) An increase in loudness and a decrease in frequency (C) A decreasing loudness and an increasing frequency (D) A decreasing loudness and a decreasing frequency (E) Just a loud thump! thump! thump!. Protons Neutrons Radiation Heat Light. 74. A radioactive isotope of iodine has a halflife of 8 hours and causes a counter to register 180 counts/min. Find the count rate for this sample of iodine two days later. (A) 1.4/min (B) 2.8/min (C) 5.6/min (D) 11.2/min (E) 22.4/min 75. The three natural radiations, in order from most penetrating to least penetrating, are (A) alpha, beta, and gamma. (B) beta, gamma, and alpha. (C) gamma, alpha, and beta. (D) gamma, beta, and alpha. (E) beta, alpha, and gamma.. 72. An atom that has lost an electron is considered to be a(n) (A) positron. (B) negatron. (C) baryon. (D) hadron. (E) ion.. STOP IF YOU FINISH BEFORE THE TEST SESSION ENDS, YOU MA Y REVIEW YOUR WORK ON THIS TEST MAY ONL Y. YOU MA Y NO T TURN TO ANY O THER TEST IN THIS BOOK. ONLY MAY NOT OTHER. Peterson’s: www.petersons.com. 27.

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(35) ANSWER SHEET. Test Code. Leave any unused answer spaces blank.. 4 2 3 6 7 8 9 1 5 O Þ O Þ O Þ O Þ O Þ O Þ O Þ O Þ O Þ 4 2 3 6 7 8 9 1 5 W O Þ O Þ O Þ O Þ O Þ O Þ O Þ O Þ O Þ 4 A 2 3 B D E 1 5 C X O Þ O Þ O Þ O Þ O Þ Y O Þ O Þ O Þ O Þ O Þ 4 2 3 6 7 8 9 1 5 O Þ O Þ O Þ O Þ O Þ O Þ O Þ O Þ O Þ. Q. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20. A B D E C O O O O O A O. B O. C O. D O. E O. A O. B O. C O. D O. E O. A B D E C O O O O O A B D E C O O O O O A B D E C O O O O O A B D E C O O O O O A B D E C O O O O O A B D E C O O O O O A B D E C O O O O O A B D E C O O O O O A O. B O. C O. D O. E O. A O. B O. C O. D O. E O. A B D E C O O O O O A B D E C O O O O O A B D E C O O O O O A B D E C O O O O O A B D E C O O O O O A B D E C O O O O O A B D E C O O O O O. Subject Test (print). V. 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40. A B D E C O O O O O A O. B O. C O. D O. E O. A O. B O. C O. D O. E O. A B D E C O O O O O A B D E C O O O O O A B D E C O O O O O A B D E C O O O O O A B D E C O O O O O A B D E C O O O O O A B D E C O O O O O A B D E C O O O O O A O. B O. C O. D O. E O. A O. B O. C O. D O. E O. A B D E C O O O O O A B D E C O O O O O A B D E C O O O O O A B D E C O O O O O A B D E C O O O O O A B D E C O O O O O A B D E C O O O O O. Peterson’s: www.petersons.com. 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60. A B D E C O O O O O A O. B O. C O. D O. E O. A O. B O. C O. D O. E O. A B D E C O O O O O A B D E C O O O O O A B D E C O O O O O A B D E C O O O O O A B D E C O O O O O A B D E C O O O O O A B D E C O O O O O A B D E C O O O O O A O. B O. C O. D O. E O. A O. B O. C O. D O. E O. A B D E C O O O O O A B D E C O O O O O A B D E C O O O O O A B D E C O O O O O A B D E C O O O O O A B D E C O O O O O A B D E C O O O O O. 29. FOR ETS USE ONLY. 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80. R/C. W/S1. A B D E C O O O O O A O. B O. C O. D O. E O. A O. B O. C O. D O. E O. A B D E C O O O O O A B D E C O O O O O A B D E C O O O O O A B D E C O O O O O A B D E C O O O O O A B D E C O O O O O A B D E C O O O O O A B D E C O O O O O A O. B O. C O. D O. E O. A O. B O. C O. D O. E O. A B D E C O O O O O A B D E C O O O O O A B D E C O O O O O A B D E C O O O O O A B D E C O O O O O A B D E C O O O O O A B D E C O O O O O. FS/S2. 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100. CS/S3. WS. A B D E C O O O O O A B D E C O O O O O A B D E C O O O O O A B D E C O O O O O A B D E C O O O O O A B D E C O O O O O A B D E C O O O O O A B D E C O O O O O A B D E C O O O O O A B D E C O O O O O A B D E C O O O O O A B D E C O O O O O A B D E C O O O O O A B D E C O O O O O A B D E C O O O O O A B D E C O O O O O A B D E C O O O O O A B D E C O O O O O A B D E C O O O O O A B D E C O O O O O.

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(37) Diagnostic Test ANSWERS AND EXPLAN ATIONS EXPLANA.

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(39) DIAGNOSTIC TEST: ANSWERS AND EXPLANATIONS. Quic k-Scor e Ans wer Quick-Scor k-Score Answ erss 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13.. C B E B B E B E A A C D C. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25. 26.. B C D B C D E E D D C B C. Peterson’s: www.petersons.com. 27. 28. 29. 30. 31. 32. 33. 34. 35. 36. 37. 38. 39.. 40. 41. 42. 43. 44. 45. 46. 47. 48. 49. 50. 51.. C C C A B A D D A A D C C. 33. C D C B D D A E B E B D. 52. 53. 54. 55. 56. 57. 58. 59. 60. 61. 62. 63.. D C B E A A B D E C B C. 64. 65. 66. 67. 68. 69. 70. 71. 72. 73. 74. 75.. E B D B A D B A E D B D.

(40) DIAGNOSTIC TEST. ANSWERS AND EXPLAN ATIONS EXPLANA. ANSWERS TO PART A. QUESTIONS 1-12 1. The cor wer is (C). The solution to the problem is found corrrect ans answ by ∆E = E3 – E2. The calculation yields +1.89 eV. The positive sign tells us the electron releases energy as it falls to a lower energy level. 2. The cor wer is (B). The solution is found again by using corrrect ans answ Bohr’s equation, ∆E = E3 – E5. The calculation yields –.966 eV. The negative sign tells us the electron absorbs energy as it moves further from the nucleus. 3. The cor corrrect ans answ (E). Also using Bohr’s equation we have wer is (E) ∆E = E3 – E1. The answer is +12.09 eV, therefore the electron releases energy. 4. The cor wer is (B) corrrect ans answ (B). The period of the pendulum is T = 2π. . We can see from the equation that the shorter the g. length of the pendulum, the shorter is the period of the pendulum. 5. The cor wer is (B). The frequency is the inverse of the answ corrrect ans period. Lengthening the pendulum increases the length of the period, which means the frequency decreases. 6. The cor wer is (E). An increase in the displacement of corrrect ans answ the pendulum raises the pendulum bob to a greater height, which gives it more potential energy. When the bob passes through the zero point, the potential energy will be transformed into greater kinetic energy, which means a greater velocity. 7. The cor wer is (B). When you set the problem into the x, corrrect ans answ y coordinate system, the negative y axis is the westward direction. Any vector in the western direction is negative.. 34. Peterson’s SAT II Success: Physics.

(41) DIAGNOSTIC TEST: ANSWERS AND EXPLANATIONS. 8. The cor wer is (E). The resultant vector is the distance corrrect ans answ the dog is from home. The opposite of the resultant vector is the equilibrant vector. 9. The cor corrrect ans answ wer is (A). The numerical value is the magnitude of the vector. The one with the largest numerical value is the longest vector. 10. The cor wer is (A) corrrect ans answ (A). The definition of the frequency of a wave is correctly stated in A 11. The cor wer is (C). This is the correct definition for corrrect ans answ wavelength. 12. The cor wer is (D) corrrect ans answ (D). The product of the frequency and the wavelength can be obtained by the dimensional analysis. V = λf , which yields  m   wave  m V =   = , a velocity unit.  wave   sec  s. ANSWERS TO PART B. QUESTIONS 13-75 13. The cor wer is (C). The coefficient of friction remains corrrect ans answ the same so the frictional force doubles, but the net force also doubles. Thus the crate is brought to a stop in the same distance. 14. The cor wer is (B). The acceleration of the skydivers corrrect ans answ continues to decrease until the terminal velocity of the group is reached. When the air resistance reaches the point where it exactly balances the force exerted on the divers by the earth’s gravitational attraction, their acceleration becomes zero. 15. The cor wer is (C). The contact between the golf club corrrect ans answ and the ball constitutes an equal and opposite force pair.. Peterson’s: www.petersons.com. 35.

(42) DIAGNOSTIC TEST. 16. The cor wer is (D). The woman pushing on the ground corrrect ans answ is one part of a force pair. The second part of the pair is the ground pushing on the woman. The force the woman exerts cannot move the ground, but the force the ground exerts does move the woman. 17. The cor wer is (B). The tension in a rope is the same corrrect ans answ throughout. Both teams must pull with a force of 1200N to maintain static equilibrium, so the total tension is 2400N. 18. The cor wer is (C). The momentum of the pitched ball is corrrect ans answ changed by the impulse Ft. The catcher reduces the magnitude of the force on his hand by increasing the time required to change the momentum of the ball. 19. The cor wer is (D). This is a conservation of momentum corrrect ans answ question. Since the momentum before an event must equal the momentum after an event, the momentum must be 12,500 kg • m/s. 20. The cor wer is (E). Use conservation of momentum for corrrect ans answ the problem. Pbefore = Pafter which is (mVf + mVp )before = (mf + mp )Vafter . The pigeon has no downward velocity before the collision, leading to: (mfVf)before = (mf + mp) Vafter Rearrange, substitute, and solve: m f Vf m f + mp. =V. 750 g • 35m / s = 22.8 m/s. 750 g + 400 g. 21. The cor wer is (E). The father holds the child at the same corrrect ans answ distance from the ground. If there is displacement, there is no work done.. 36. Peterson’s SAT II Success: Physics.

(43) DIAGNOSTIC TEST: ANSWERS AND EXPLANATIONS. 22. The cor wer is (D). The larger amount of padding in the corrrect ans answ amateur gloves causes the time of impact to be increased, which results in less force being exerted. 23. The cor corrrect ans answ wer is (D). The same car traveling at twice the velocity has four times the energy. Constant frictional force from the tires will therefore require four times the stopping distance. The car moving at the lower rate only requires. 1 the stopping 4. distance. 24. The cor corrrect ans answ wer is (C). This is a work-energy problem. PE = Work → mgh = F•D. Rearrange, substitute, and solve (remember m • g = wt): mgh 19.6 N • 2 m = F∴ = 1600 N D .0245m. 25. The cor wer is (B). The velocity of the pile driver at the corrrect ans answ 2 time it first contacts the head of the nail is Vf = 2gs. Since, by. 1 mVp2 = mgd. V is also the original 2 velocity as the nail is driven into the wood for a distance of .0245m. conservation of energy,. Stating the equations: V f = 2 gd 2. and V0 = 2 − ad 2. but V f = V0 2. and. 2. 2 • 9.8m/s2 • 2m = −800m/s2 2 • −.0245m. 26. The cor wer is (C). All the gravitational potential energy corrrect ans answ at the top of the child’s swing pathway is converted into kinetic energy at the bottom of her swing pathway. 27. The cor wer is (C). The wheel is moving at a constant corrrect ans answ rate (angular velocity). There is no angular acceleration because ∆ω is zero.. Peterson’s: www.petersons.com. 37.

(44) DIAGNOSTIC TEST. 28. The cor wer is (C). The diver needs to complete the corrrect ans answ rotations as quickly as possible. The tuck position reduces the moment of inertia of the diver’s body, allowing the diver to spin at a higher rate. 29. The cor wer is (C). Regardless of where on the merrycorrrect ans answ go-round you are located during the ride, your angular displacement is the same for the whole ride. 30. The cor wer is (A). The central axis would have a gravitacorrrect ans answ tional force of zero because there is no radial distance to provide a centripetal acceleration or centripetal force. 31. The cor wer is (B). Although the person has moved corrrect ans answ further from the center of the earth, that person now has the additional mass of Mt. Everest under him.. m1 • m2 , the person will r2 be weightless at the center of the earth because r = 0, and all the mass of the earth is attracting the person from outside the center.. 32. The cor wer is (A). Since F = G corrrect ans answ. wer is (D). The apparent weight is zero. The 33. The cor corrrect ans answ person is in free fall. No forces are acting on the person except gravity. 34. The cor wer is (D). The centripetal force required to corrrect ans answ keep a satellite in orbit remains constant. That means the closer to earth a satellite is, the faster it must travel in its orbit mV 2 . The farther the satellite is located from the earth, the r slower it travels in orbit. Fc =. 35. The cor corrrect ans answ wer is (A). The geosynchronous satellite circles the earth at the same rate as the earth turns beneath it. Thus the satellite stays in the same position relative the earth. 36. The cor wer is (A). Most of the alpha particles passed corrrect ans answ through the gold foil, leading Rutherford to conclude that atoms were composed mostly of space with a dense core.. 38. Peterson’s SAT II Success: Physics.

(45) DIAGNOSTIC TEST: ANSWERS AND EXPLANATIONS. 37. The cor wer is (D). Based on Hooke’s Law, F = kx corrrect ans answ where F is mg and x is the elongation. Solving for x =. mg , so if m k. is doubled, x is doubled as well.. wer is (C). The same number of gas particles 38. The cor corrrect ans answ. m v where mass can be restated as the mass of the gas = (mol. Mass) (Avogadro’s Number). Thus the mass is directly related to the number of particles in the space. More particles in less space yield greater density. now occupies half the volume. The formula for density is d =. 39. The cor wer is (C). Refrigerators and freezers operate in answ corrrect ans a reversed direction from a heat engine. Instead of using a hot source to produce work, work is done on a warm source and heat is removed, producing the cooling effect. 40. The cor wer is (C). The water vapor in the can is hot corrrect ans answ and at atmospheric pressure. Cooling the can quickly condenses the water vapor, which also reduces the pressure inside the can. The outside atmospheric pressure is greater, thus crushing the can. 41. The cor wer is (D). The first law of thermodynamics can corrrect ans answ be stated as:.  the heat added   the work done   the increase in   to a system  −  by the system  =  system energy        or ∆Q − ∆W = ∆U. Peterson’s: www.petersons.com. 39.

(46) DIAGNOSTIC TEST. 42. The cor wer is (C). This is a calorimetric problem. Since corrrect ans answ the ratio of C° to K is 1 to 1, we will not change temperature scales. Our working equation is Cm1∆ Τhot + Cm2 ∆ Τcold = 0 Cm1 (t f − t0 )hot + Cm2 (t f − t0 )cold = 0 factor out the C ’s m(t f − t0 ) + m(t f − to ) = 0 360 g(t f − 95°C ) + 275g(t f − 10°C ) = 0 360 gt f − 34200 gC ° + 275ggt f − 2750 gC ° = 0 635gt f − 36950 gC ° = 0 36950 gC ° tf = = 58°C 635g 43. The cor wer is (B). The velocity of a wave is v = λ • f . corrrect ans answ Substitute into the equation and solve. v =λ• f  meters   waves  =  16   .5  wave   second   meters =8 second. 44. The cor wer is (D). If the input frequency of a wave corrrect ans answ system is equal to its vibrating frequency, the resonant system acts to amplify the waves, thus transferring their energy into the delicate crystal and breaking it. 45. The cor corrrect ans answ wer is (D). When the two impulses are located at the same point they will algebraically add the two positive peaks, yielding a larger wave, which is equal to both of their magnitudes. 46. The cor wer is (A). The two waves are moving away corrrect ans answ from one another. 47. The cor wer is (E). When the two pulses are located at corrrect ans answ the same point they will algebraically add the positive peak to the negative peak yielding no wave or disturbance. They have destroyed one another at that spot.. 40. Peterson’s SAT II Success: Physics.

(47) DIAGNOSTIC TEST: ANSWERS AND EXPLANATIONS. 48. The cor wer is (B). The sound waves from the approachcorrrect ans answ ing whistle strike the listener’s ears at an increasing rate, producing the increasing pitch. wer is (E). The charge on the two does not 49. The cor corrrect ans answ change, so the force between them MUST operate in the same direction. The magnitude of the force increases by 4. 50. The cor wer is (B). Charge will distribute itself equally corrrect ans answ over the surface of a body because of the repulsion the electrons have for one another.. 51. The cor wer is (D). Coulombs Law is F = k corrrect ans answ. q1q2 . r2. −6 −6  m 2  (−3 × 10 f )(−4 × 10 f ) = 270 N The solution is F =  9 × 10 9 N 2  (.02 m)2 C  . 52. The cor wer is (D). The electric field between two corrrect ans answ parallel plates is constant throughout. The value of the field is defined as E =. F where E is the electric field, F is force, and q is q. charge. wer is (C). The value of the electric field about a 53. The cor corrrect ans answ. q . The distance from the point in question r2 determines the strength of the electric field in this case. point source is E = k. 54. The cor wer is (B). This is a problem that requires an answ corrrect ans Ohm’s Law calculation to find resistance. Rearrange the equation to find resistance.. V = IR R=. Peterson’s: www.petersons.com. 41. V 6V = = 4Ω I 1.5 A.

(48) DIAGNOSTIC TEST. 55. The cor wer is (E). The higher internal resistance in the corrrect ans answ battery leads to a reduction in the terminal potential of the battery. The lower terminal potential can produce less current in the circuit. 56. The cor wer is (A). The total resistance is found by using corrrect ans answ the reciprocal method. Rt =. 1 1 1 + R1 R2. =. 1 1 1 + 3Ω 6Ω. =. 1 = 2Ω 1 2. 57. The cor wer is (A). V = IR, rearrange and I = corrrect ans answ. I=. V Rt. 6V 6V = = 1A 2Ω + 2Ω + 2Ω 6Ω. 58. The cor wer is (B). Use Ohm’s Law to solve the problem. corrrect ans answ. V = IR (1A)(2Ω) = 2V 59. The cor wer is (D). The more magnetic field lines a corrrect ans answ charged particle crosses, the stronger the force exerted on the particle. Moving through the magnetic field in a perpendicular path (90°) will cause the particle to cross the most field lines. wer is (E). Your computer picture is caused by 60. The cor answ corrrect ans manipulating a magnetic yoke to produce a picture. The interference with radio is caused by magnetic lines and radio waves interacting. VHS films are produced by magnetic pulses, and the Aurora Borealis is the visual show produced when cosmic particles enter the earth’s magnetic field. 61. The cor wer is (C). Motors are used in a variety of corrrect ans answ mechanical devices such as saws, sanders, drills, and pumps. The electric energy is converted to mechanical energy by the rotation of a loop of wire in a magnetic field.. 42. Peterson’s SAT II Success: Physics.

(49) DIAGNOSTIC TEST: ANSWERS AND EXPLANATIONS. 62. The cor wer is (B). The transformer equation is corrrect ans answ Vp Vs. =. Vs = Vs =. # of primary turns #of secondary turns (Vp )(#of secondary turns) # of primary turns (240V )(500 turns) = 30V 4000 turns. 63. The cor wer is (C). Your image is standing upright. Your corrrect ans answ feet and the image of your feet are down, and your head and the image of your head are up. When you raise your right hand, the left hand of the image is raised. This is a reversal of left and right. wer is (E). Light slows down when it enters a 64. The cor corrrect ans answ material that is more optically dense. Since the velocity is related to both the frequency and the wavelength by V = λf it follows that one of these quantities will change, too. Since the distance between the waves changes at the optical boundary, the wavelength also changes. The frequency remains unchanged. 65. The cor wer is (B) answ (B). A ray parallel to the central radius is corrrect ans reflected through the focal point. 66. The cor wer is (D). The relative retardation of the light corrrect ans answ wave (the letter D in the diagram) is also its wavelength. The bright spots produced on the screen represent n = 1,2,3, etc., whole number values of wavelengths. 67. The cor wer is (B). The zeroth fringe is the central bright corrrect ans answ spot where both waves from the two slits have traveled the same distance. 68. The cor wer is (A). The wave nature of light does not corrrect ans answ satisfactorily explain why low frequency light cannot energize electrons from the surface of metals, which high frequency light does. The intensity of the light doesn’t matter. This is explained by the particle theory in which photons carry energy, which is transferred to electrons in the metal.. Peterson’s: www.petersons.com. 43.

(50) DIAGNOSTIC TEST. 69. The cor wer is (D). The energized atoms of the gas emit corrrect ans answ energy in specific regions of the visible spectrum called line spectra. 70. The cor corrrect ans answ wer is (B). One of two postulates of Einstein’s theory of special relativity is that the speed of light in a vacuum is the same for all observers. 71. The cor wer is (A). A combination of Doppler effect and corrrect ans answ matter waves. The waves being emitted by the moving fullback reach you at a rate greater than they are produced. Additionally the approaching fullback is getting louder, too. 72. The cor wer is (E). All charged particles are called ions. corrrect ans answ 73. The cor wer is (D). The reactors produce heat, which in corrrect ans answ turn changes water to steam, which is used to turn the turbines to generate electricity. 74. The cor wer is (B). Half-life means that half of the corrrect ans answ substance decays away during that time. Over 48 hours there are 6 half-lives.. 48 hours. 8 hours = 6 half lives halflife. 6. 1   (180 counts/min) = 2.8 counts/min 2. 75. The cor wer is (D). The penetrating capability of radiacorrrect ans answ tion is based upon its energy. Gamma radiation is most energetic followed by beta then alpha.. 44. Peterson’s SAT II Success: Physics.

(51) Chapter 1 PRELIMIN AR Y CONCEPTS PRELIMINAR ARY.

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(53) CHAPTER 1 PRELIMIN AR Y CONCEPTS PRELIMINAR ARY SIMPLE EQUA TIONS AND ALGEBRA EQUATIONS Physics is the branch of science that studies physical phenomena. Why does the sound of a starter’s pistol reach an observer standing at the race finish line after the puff of smoke is seen when the gun is fired? Why do some objects float in water, while other objects sink? What causes a ball to roll downhill? These questions and more are the stuff of physics. Sometimes the study of physical phenomena involves observation, experimentation, and calculation. The calculations used in this review are accomplished by using a little algebra, a little geometry, and a little trigonometry. Many relationships in physics can be expressed as equations. For example: F = ma s = vt Sometimes the value to be found is not by itself (isolated) in the equation. Take, for example, the linear motion equation: 2. 2. Vf = V0 + 2as Let’s suppose the acceleration a is the only unknown quantity in the equation. We must perform a little algebraic manipulation to isolate a so that it is the only quantity on its side of the equal sign. Before starting the process, remember one simple rule: whatever operation you perform on one side of the equal sign, you must also perform on the other side of the equal sign. Let’s begin by isolating a in the equation below: Vf 2 = V0 2 + 2as 2. We then subtract Vo from both sides: Vf2 –Vo2 = Vo2 –Vo2 + 2as. Peterson’s: www.petersons.com. 47.

(54) CHAPTER 1. This leads to: Vf2 – Vo2 = 2as. Then we divide both sides by 2s: V f − V02 2. 2s. =. 2 as 2s. Clearing fractions gives: V f 2 − V02 2s. =a. If you find this algebra to be difficult, you might need to stop here and review your math skills before proceeding.. GRAPHS Some relationships studied in physics exist between experimental values. The relationships between these quantities can be shown using a technique called graphing aphing. With this technique, one quantity (the independent variable) is carefully controlled, while the other quantity (the dependent variable) is measured at each value of the independent variable. The independent variable is plotted along the x axis of the graph, and the dependent variable is plotted along the y axis of the graph. The slope for the graph is calculated by dividing the ∆Y by the ∆X .. ∆y ∆x A graph is a picture of the relationship between two or more quantities. A direct relationship is one in which both quantities increase (or decrease) in the same manner. In an inverse relationship, one quantity increases and the other decreases. A parabolic relationship exists when one quantity varies as the square of the other. Slope =. 48. Peterson’s SAT II Success: Physics.

(55) GRAPHS. DIRECT RELATIONSHIP. GRAPH A Graph A shows a linear relationship between x and y. For every 2 units’ increase in variable x, variable y increases by 1 unit. The slope of the graph is calculated as rise over run.. Peterson’s: www.petersons.com. 49.

(56) CHAPTER 1. INVERSE RELATIONSHIP. GRAPH B Graph B shows an indirect linear relationship between x and y.. 50. Peterson’s SAT II Success: Physics.

(57) GRAPHS. PARABOLIC RELATIONSHIP. GRAPH C Graph C shows a parabolic relationship between x and y. The ability to read a graph is crucial. They are used to display and compare physical concepts. Be sure you can read and evaluate graphs.. Peterson’s: www.petersons.com. 51.

(58) CHAPTER 1. RIGHT TRIANGLES Any triangle with a 90° internal angle is defined as a right triangle. In such a triangle, the remaining two internal angles of the right triangle sum to an additional 90°. If you are sure you are dealing with a right triangle, the Pythagorean theorem becomes a useful tool when trying to find one of the sides.. The right triangle above is labeled according to convention. The side opposite the 90° angle is called the hypotenuse and is labeled c. The side beside the smaller of the remaining two angles is called the adjacent side b, and the side across from the smaller angle is called the opposite side a. The Pythagorean theorem is stated mathematically as 2 c = a2 + b2. Thus, should we know the size of any two sides, we can find the missing side.. Side c = a 2 + b 2 Side a = c 2 − b 2 Side b = c 2 − a 2 Right triangles have special relationships not only with their sides but also with their internal angles. For every right triangle, the sides of that triangle and the internal angles of that triangle form a ratio.. 52. Peterson’s SAT II Success: Physics.

(59) RIGHT TRIANGLES. These are given as: sine θ =. opposite hypotenuse. or sin θ =. a c. cosine θ =. adjacent hypoteenuse. or cos θ =. b c. opposite adjacent. or tan θ =. a b. tangent θ =. Further, no matter how long the sides of the triangle may be, the ratio between the three sides remains the same as long as the internal angle remains the same.. Triangles 1, 2, 3, and 4 above all have an internal angle of 30°. Thus the sine, cosine, and tangent values are the same in all four triangles.. CARTESIAN COORDINATES The construction and use of right triangles is a valuable tool in solving physics problems. Unfortunately, not all measures and quantities begin neatly at zero. Rather than allow this to complicate matters, we simplify by using the Cartesian Coordinate System in conjunction with triangles. You may also recognize this as the four quadrant x and y system.. Peterson’s: www.petersons.com. 53.

(60) CHAPTER 1. • The region between the x and y coordinates on the upper right is called the first quadrant. All x and y quantities in the first quadrant are positive. • The region between the x and y coordinates on the upper left is called the second quadrant. All x quantities in the second quadrant are negative, and all y quantities in the second quadrant are positive. • The region between the x and y coordinates on the lower left is called the third quadrant. All x and y quantities in the third quadrant are negative. • The region between the x and y coordinates on the lower right is called the fourth quadrant. All x quantities in the fourth quadrant are positive and all y quantities in the fourth quadrant are negative.. X coordinate +x --x --x +x. Quadrant I II III IV. 54. Y coordinate +y +y --y --y. Peterson’s SAT II Success: Physics.

(61) RIGHT TRIANGLES. Situations where the Cartesian Coordinate System is used conventionally follow the x, y system for labeling purposes. When a right triangle is placed within the coordinate system, its sides can be renamed as follows:. Thus, we could look at the trigonometry functions for a right triangle that has been placed into the coordinate system as: sin θ =. a c. changes to →. sin θ =. y r. cos θ =. b c. changes to →. cos θ =. x r. tan θ =. a b. changes to →. cos θ =. y x. The conventional a, b, c labeling is changed to represent the physics application, thus becoming x, y, r, where r is the hypotenuse. The Pythagorean Theorem now looks like this:. Conventional labeling becomes coordinate system labeling.. Peterson’s: www.petersons.com. 55.

(62) CHAPTER 1. UNITS AND CONVERSIONS Oftentimes students will be given a value for a physical quantity such as 50 km/hr and asked to perform operations that require them to convert to m/sec. Finding the correct units in this situation means that you must be able to change kilometers to meters and to change hours to seconds. Experience has shown most students to be very capable at performing the mathematic operation. The problem seems to be one of setting up the conversions between the differing units. 50 km/hr = ? m/sec One factor that can cause difficulty with conversions is that many students lack some basic information. You should know instantly that there are 60 seconds in one minute and 3600 seconds in one hour. The metric units you should know are: 1000 meters = 1 kilometer 100 centimeters = 1 meter 1000 millimeters = 1 meter Then there are the equivalent values for volume (liters) and mass (grams) as well. There is a list of constants, equivalents, and physics formulas on page 15. Be sure you are very familiar with this information. Returning to the conversion mentioned, we begin: 50 km/hr = ? m/sec There are 1000 meters in a kilometer, thus we multiply 50 km by 1000 m/km, yielding 50,000 meters. Now we have: 50,000 m/hr = ? m/sec There are 3600 seconds per hour, thus we multiply one hour by 3600 sec/hr, yielding 3600 seconds. Divide distance by time for the final value. 50, 000m = 13.88m / sec 3600 sec 56. Peterson’s SAT II Success: Physics.

(63) SCALARS AND VECTORS. The whole conversion can be made into a one-step calculation by writing each part and solving.. 1   (50 km/hr)(1,000 m/km)  3600sec  = 13.88m/sec   hr   During the conversion process it is absolutely essential that you state and complete your dimensional analysis.. SCALARS AND VECTORS. SCALARS The treatment of physical quantities is based upon the kind of quantity in question. Two important physical quantities we will study are scalars and vectors. Quantities defined strictly by their magnitude are called scalar quantities. Scalars are easy to recognize: a dozen eggs, a gross of paper clips, half a dozen apples, a kilogram of cheese. All the stated quantities denote a number (12, 144, 6, 1) of the item listed, nothing more. Therefore, a Sunday drive could easily be described by the number of miles traveled: “We drove 80 km today.” Should we wish to add more information, such as where we traveled, then another physical quantity is used.. VECTORS Quantities defined both by their magnitude and direction are called vector quantities. Vectors are also easily recognized, as they must include both magnitude and direction: four steps to the right, 50 meters south, 9N @ 45º. Taking another look at the Sunday drive above, we can define it as a vector by saying “We drove 80 km to the museum and back today.” Vectors are always stated in respect to a reference point.. Peterson’s: www.petersons.com. 57.

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